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Analytical Geometry examples

Example 1 of Ellipse Equation

The standard form of an ellipse is:
[(x - h)2)/a2 + [(y - k)2)/b2 = 1
the center of the ellipse is (h, k)
and its graph is:
the graph of the stadand ellipse with center is (h, k).

Example 2 of Ellipse Equation

Question 1.
Find the center of the ellipse given by the equation 9x2 - 36x + 4y2 + 24y + 36 = 0 and draw the ellipse.
Solution:
change the given ellipse equation into the stand ellipse equation
9x2 - 36x + 4y2 + 24y + 36 = 0
9(x2 - 4x) + 4(y2 + 6y) + 36 = 0
9[x2 - 4x + (4/2)2 - (4/2)2] + 4[y2 + 6y + (6/2)2 - (6/2)2] + 36 = 0
9[x2 - 4x + 4 - 4] + 4[y2 + 6y + 9 - 9] + 36 = 0
9[x2 - 4x + 4] - 9 × 4 + 4[y2 + 6y + 9] - 4 × 9 + 36 = 0
9[x2 - 4x + 22] - 36 + 4[y2 + 6y + 32] - 36 + 36 = 0
9[x2 - 4x + 22] + 4[y2 + 6y + 32] = 36
both side of the equation divide by 36 (each item divide by 36)
9[x2 - 4x + 22]/36 + 4[y2 + 6y + 32]/36 = 36/36
[x2 - 4x + 22]/4 + [y2 + 6y + 32]/9 = 1
[(x - 2)2]/22 + [(y + 3)2]/32 = 1
the standard ellipse equation is: [(x - 2)2]/22 + [(y + 3)2]/32 = 1
therefore, the center of the given ellipse is: (2 , - 3)
The graphic of the ellipse is
graphic of ellipse

Example 3 of Circle Equation

Question 2
Find the center of the circle x2 - 4x + y2 + 2y - 4 = 0, draw the graph and find its area.
Solution
x2 - 4x + y2 + 2y - 4 = 0
x2 - 4x + (4/2)2 - (4/2)2 + y2 + 2y + (2/2)2 - (2/2)2 - 4 = 0
x2 - 4x + (2)2 - (2)2 + y2 + 2y + (1)2 - (1)2 - 4 = 0
x2 - 4x + 4 - 4 + y2 + 2y + 1 - 1 - 4 = 0
x2 - 4x + 4 + y2 + 2y + 1 - 4 - 1 - 4 = 0
(x2 - 4x + 4) + (y2 + 2y + 1) - 4 - 1 - 4 = 0
(x2 - 4x + 4) + (y2 + 2y + 1) = 4 + 1 + 4
(x2 - 4x + 4) + (y2 + 2y + 1) = 9
(x - 2)2 + (y + 1)2 = 32
therefore, the center of the circle is: (2, -1) and the radius of the circle is: 3
Find the Circle Equation
The center of the circle is (2, -1), and the radius of the circle is r = 3
The area of the circle is pi*r2 = 3.14 * 9 = 28.26 square units.
remark:
circle equation: (x - xo)2 + (y - yo)2 = r2
center of the circle is: (xo , yo) and radius is r